CIVL 1112
Surveying - Traverse Calculations
1/13
Surveying - Traverse
Surveying - Traverse
Surveying - Traverse
Distance - Traverse
Introduction
Methods of Computing Area
 Almost all surveying requires some calculations to
reduce measurements into a more useful form for
determining distance, earthwork volumes, land areas,
etc.
 A simple method that is useful for rough area
estimates is a graphical method
 A traverse is developed by measuring the distance
and angles between points that found the boundary of
a site
 We will learn several different techniques to compute
the area inside a traverse
 In this method, the
traverse is plotted to
scale on graph paper,
and the number of
squares inside the
traverse are counted
B
A
C
D
Distance - Traverse
Methods of Computing Area
B
a
A
1
Area ABC  ac sin 
2
b

c
C
Distance - Traverse
Methods of Computing Area
B
a
A
Area ABD 
b


d
c
C
1
ad sin 
2
1
2
Area BCD  bc sin 
D
Area ABCD  Area ABD  Area BCD
CIVL 1112
Surveying - Traverse Calculations
Distance - Traverse
Surveying - Traverse
Methods of Computing Area
B
b
A
Area ABE 
c


e
Balancing Angles
C
a
1
ae sin 
2
 Before the areas of a piece of land can be computed,
it is necessary to have a closed traverse
 The interior angles of a closed traverse should total:
1
2
D
(n - 2)(180°)
Area CDE  cd sin 
d
2/13
E
where n is the number of sides of the traverse
 To compute Area BCD more data is required
Surveying - Traverse
Surveying - Traverse
Balancing Angles
Balancing Angles
A
Error of closure
B
D
 A surveying heuristic is that the total angle should not
vary from the correct value by more than the square
root of the number of angles measured times the
precision of the instrument
 For example an eight-sided traverse using a 1’ transit,
the maximum error is:
1' 8  2.83'  3'
C
Angle containing mistake
Surveying - Traverse
Surveying - Traverse
Balancing Angles
Latitudes and Departures
 If the angles do not close by a reasonable amount,
mistakes in measuring have been made
 The closure of a traverse is checked by computing
the latitudes and departures of each of it sides
 If an error of 1’ is made, the surveyor may correct
one angle by 1’
 If an error of 2’ is made, the surveyor may correct
two angles by 1’ each
 If an error of 3’ is made in a 12 sided traverse, the
surveyor may correct each angle by 3’/12 or 15”
N
N
B
Latitude AB
E
W
Bearing 
A
S
Departure AB
Bearing 
W
C
Latitude CD
S
Departure CD
D
E
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse
3/13
Surveying - Traverse
Latitudes and Departures
Error of Closure
 The latitude of a line is its projection on the north–
south meridian
 Consider the following statement:
N
Latitude AB
E
W
Bearing 
A
“If start at one corner of a closed traverse and walk its
lines until you return to your starting point, you will have
walked as far north as you walked south and as far east as
you have walked west”
 The departure of a line is
its projection on the east–
west line
B
Departure AB
 A northeasterly bearing
has a + latitude and
+ departure
 Therefore --
 latitudes = 0
and
 departures = 0
S
Surveying - Traverse
Surveying - Traverse
Error of Closure
Error of Closure
 When latitudes are added together, the resulting
error is called the error in latitudes ( EL )
 If the measured bearings and distances are plotted
on a sheet of paper, the figure will not close because
of EL and ED
 The error resulting from adding departures together
is called the error in departures ( ED )
B
EL
A
Surveying - Traverse
234.58’
Eclosure
perimeter
Departure AB
A
W
175.18’
S 6° 15’ W
C
B
W  (189.53 ft ) sin(615')  20.63 ft
E
Latitude AB
189.53’
S  (189.53 ft ) cos(615')  188.40 ft
175.18’
D N 81° 18’ W
2
N
S 29° 38’ E
142.39’
 ED 
Latitudes and Departures - Example
S 6° 15’ W
B
N 12° 24’ W
Precision 
2
Surveying - Traverse
189.53’
E
E L 
 Typical precision: 1/5,000 for rural land, 1/7,500
for suburban land, and 1/10,000 for urban land
A
N 42° 59’ E
Eclosure 
C
D
Latitudes and Departures - Example
Error of closure
ED
S
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse
Surveying - Traverse
Latitudes and Departures - Example
Latitudes and Departures - Example
Bearing
Side
N
Departure BC
E  (175.18 ft ) sin(2938')  86.62 ft
B
W
4/13
E
175.18’
AB
BC
CD
DE
EA
degree
m inutes
6
29
81
12
42
15
38
18
24
59
S
S
N
N
N
Length (ft)
Latitude
Departure
189.53
175.18
197.78
142.39
234.58
939.46
-188.403
-152.268
29.916
139.068
171.607
-0.079
-20.634
86.617
-195.504
-30.576
159.933
-0.163
W
E
W
W
E
Latitude BC
S 29° 38’ E
S  (175.18 ft ) cos(2938')  152.27 ft
C
S
Surveying - Traverse
Surveying - Traverse
Latitudes and Departures - Example
Bearing
Side
AB
BC
CD
DE
EA
Eclosure 
S
S
N
N
N
E L 
2
degree
m inutes
6
29
81
12
42
15
38
18
24
59
 ED  
2
Group Example Problem 1
Length (ft)
Latitude
Departure
189.53
175.18
197.78
142.39
234.58
939.46
-188.403
-152.268
29.916
139.068
171.607
-0.079
-20.634
86.617
-195.504
-30.576
159.933
-0.163
W
E
W
W
E
2
 0.079 
Eclosure
2
  0.163
0.182 ft
Precision 


perimeter 939.46 ft
 0.182 ft
1
5,176
A
S 77° 10’ E
N 29° 16’ E
651.2’
B
660.5’
S 38° 43’ W
D
491.0’
826.7’
N 64° 09’ W
C
Surveying - Traverse
Surveying - Traverse
Balancing Latitudes and Departures
Group Example Problem 1
 Balancing the latitudes and departures of a traverse
attempts to obtain more probable values for the
locations of the corners of the traverse
Bearing
Side
AB
BC
CD
DE
S
S
N
N
degree
m inutes
77
38
64
29
10
43
9
16
Length (ft)
E
W
W
E
651.2
826.7
491.0
660.5
Latitude
Departure
 A popular method for balancing errors is called the
compass or the Bowditch rule
 The ‘Bowditch rule’ was devised by Nathaniel
Bowditch, surveyor, navigator and mathematician,
as a proposed solution to the problem of compass
traverse adjustment, which was posed in the
American journal The Analyst in 1807.
CIVL 1112
Surveying - Traverse Calculations
5/13
Surveying - Traverse
Surveying - Traverse
Balancing Latitudes and Departures
Balancing Latitudes and Departures
A
 The compass method assumes:
N 42° 59’ E
1) angles and distances have same error
2) errors are accidental
S 6° 15’ W
234.58’
189.53’
B
 The rule states:
E
“The error in latitude (departure) of a line is to
the total error in latitude (departure) as the
length of the line is the perimeter of the traverse”
S 29° 38’ E
142.39’
N 12° 24’ W
175.18’
175.18’
D N 81° 18’ W
Surveying - Traverse
C
Surveying - Traverse
Latitudes and Departures - Example
Latitudes and Departures - Example
 Recall the results of our example problem
 Recall the results of our example problem
Side
Length (ft)
Bearing
AB
BC
CD
DE
EA
S
S
N
N
N
degree
m inutes
6
29
81
12
42
15
38
18
24
59
W
E
W
W
E
Latitude
Departure
Side
189.53
175.18
197.78
142.39
234.58
Bearing
AB
BC
CD
DE
EA
S
S
N
N
N
degree
m inutes
6
29
81
12
42
15
38
18
24
59
W
E
W
W
E
Length (ft)
Latitude
Departure
189.53
175.18
197.78
142.39
234.58
939.46
-188.403
-152.268
29.916
139.068
171.607
-0.079
-20.634
86.617
-195.504
-30.576
159.933
-0.163
Surveying - Traverse
Surveying - Traverse
Balancing Latitudes and Departures
Balancing Latitudes and Departures
N
N
Latitude AB
Departure AB
S  (189.53 ft ) cos(6 15')  188.40 ft
W  (189.53 ft ) sin(615')  20.63 ft

A
W
E
189.53’
B
Correction in LatAB
EL
S 6° 15’ W

Correction in LatAB 
LAB
perimeter
EL  LAB 
perimeter
S
Correction in LatAB 
0.079 ft 189.53 ft 
939.46 ft
 0.016 ft
A
W
E
189.53’
B
Correction in DepAB
ED
S 6° 15’ W

Correction in DepAB 
LAB
perimeter
ED  LAB 
perimeter
S
Correction in DepAB 
0.163 ft 189.53 ft 
939.46 ft
 0.033 ft
CIVL 1112
Surveying - Traverse Calculations
6/13
Surveying - Traverse
Surveying - Traverse
Balancing Latitudes and Departures
Balancing Latitudes and Departures
N
N
Latitude BC
Departure BC
S  (175.18 ft ) cos(29 38')  152.27 ft

B
W
E
Correction in LatBC
EL
175.18’
S 29° 38’ E

Correction in LatBC 
C
S
Correction in LatBC 
LBC
E  (175.18 ft ) sin(2938')  86.62 ft
B
W
E
perimeter
939.46 ft
ED
175.18’
EL  LBC 
S 29° 38’ E
perimeter
0.079 ft 175.18 ft 
Correction in DepBC
perimeter
ED  LBC 
Correction in DepBC 
C
S
 0.015 ft
LBC

Correction in DepBC 
perimeter
0.163 ft 175.18 ft 
939.46 ft
Surveying - Traverse
Surveying - Traverse
Balancing Latitudes and Departures
Balancing Latitudes and Departures
Length (ft)
189.53
175.18
197.78
142.39
234.58
939.46
Latitude
-188.403
-152.268
29.916
139.068
171.607
-0.079
Departure
-20.634
86.617
-195.504
-30.576
159.933
-0.163
Corrections
Latitude Departure
0.016
0.015
Balanced
Latitude Departure
0.033
0.030
Length (ft)
189.53
175.18
197.78
142.39
234.58
939.46
Latitude
-188.403
-152.268
29.916
139.068
171.607
-0.079
Departure
Corrections
Latitude Departure
-20.634
86.617
-195.504
-30.576
159.933
-0.163
0.016
0.015
Balanced
Latitude Departure
0.033
0.030
-188.388
-152.253
Surveying - Traverse
Surveying - Traverse
Balancing Latitudes and Departures
Balancing Latitudes and Departures
189.53
175.18
197.78
142.39
234.58
939.46
Latitude
-188.403
-152.268
29.916
139.068
171.607
-0.079
Departure
-20.634
86.617
-195.504
-30.576
159.933
-0.163
Corrections
Latitude Departure
0.016
0.015
0.017
0.012
0.020
-20.601
86.648
Corrected latitudes and departures
Corrections computed on previous slides
Length (ft)
 0.030 ft
0.033
0.030
0.034
0.025
0.041
Balanced
Latitude Departure
-188.388
-152.253
29.933
139.080
171.627
0.000
No error in corrected latitudes and departures
-20.601
86.648
-195.470
-30.551
159.974
0.000
 Combining the latitude and departure calculations with
corrections gives:
Side
AB S
BC S
CD N
DE N
EA N
Bearing
degree
minutes
6
29
81
12
42
15
38
18
24
59
Length (ft) Latitude
W
E
W
W
E
189.53
175.18
197.78
142.39
234.58
939.46
-188.403
-152.268
29.916
139.068
171.607
-0.079
Departure
-20.634
86.617
-195.504
-30.576
159.933
-0.163
Corrections
Latitude Departure
0.016
0.015
0.017
0.012
0.020
0.033
0.030
0.034
0.025
0.041
Balanced
Latitude
Departure
-188.388
-152.253
29.933
139.080
171.627
0.000
-20.601
86.648
-195.470
-30.551
159.974
0.000
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse
7/13
Surveying - Traverse
Group Example Problem 2
Group Example Problem 3
 Balance the latitudes and departures for the following
traverse.
 In the survey of your assign site in Project #3, you will
have to balance data collected in the following form:
Corrections
Balanced
Length (ft) Latitude Departure Latitude Departure Latitude Departure
N
600.0
450.0
750.0
1800.0
450.00
-285.00
-164.46
0.54
339.00
259.50
-599.22
-0.72
B
N 69° 53’ E
A
51° 23’
713.93’
105° 39’
781.18’
606.06’
78° 11’
124° 47’
D
Surveying - Traverse
Surveying - Traverse
Calculating Traverse Area
Group Example Problem 3
 In the survey of your assign site in Project #3, you will
have to balance data collected in the following form:
Corrections
Length (ft) Latitude Departure Latitude Departure
Bearing
Side
Balanced
Latitude
Departure
degree m inutes
AB N
BC
CD
DA
69
53
E
713.93
606.06
391.27
781.18
Eclosure =
Precision =
 The best–known procedure for calculating land areas
is the double meridian distance (DMD) method
 The meridian distance of a line is the east–west
distance from the midpoint of the line to the
reference meridian
 The meridian distance is positive (+) to the east and
negative (-) to the west
ft
1
Surveying - Traverse
Calculating Traverse Area
N
A
234.58’
S 6° 15’ W
189.53’
B
E
S 29° 38’ E
142.39’
N 12° 24’ W
Surveying - Traverse
Calculating Traverse Area
N 42° 59’ E
Reference
Meridian
C
391.27’
175.18’
175.18’
D N 81° 18’ W
C
 The most westerly and easterly points of a traverse
may be found using the departures of the traverse
 Begin by establishing a arbitrary reference line and
using the departure values of each point in the
traverse to determine the far westerly point
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse
Surveying - Traverse
Calculating Traverse Area
Length (ft)
189.53
175.18
197.78
142.39
234.58
939.46
Latitude
Departure
-188.403
-152.268
29.916
139.068
171.607
-0.079
-20.634
86.617
-195.504
-30.576
159.933
-0.163
-20.601
D
-30.551
E
B
-195.470
D
0.033
0.030
0.034
0.025
0.041
A
B
159.974
E
Calculating Traverse Area
Corrections
Latitude Departure
0.016
0.015
0.017
0.012
0.020
-188.388
-152.253
29.933
139.080
171.627
0.000
-20.601
86.648
-195.470
-30.551
159.974
0.000
Reference
Meridian
D
189.53’
S 29° 38’ E
142.39’
175.18’
175.18’
Point E is the
farthest to the west
D N 81° 18’ W
C
Surveying - Traverse
DMD Calculations
Meridian distance
of line AB
 The meridian distance of line
AB is equal to:
the meridian distance of EA
+ ½ the departure of line EA
+ ½ departure of AB
A
A
C
S 6° 15’ W
234.58’
N 12° 24’ W
N
E
N 42° 59’ E
B
N
B
A
E
 The meridian distance
of line EA is:
A
Reference
Meridian
C
C
Surveying - Traverse
N
N
Balanced
Latitude Departure
86.648
A
DMD Calculations
8/13
B
E
 The DMD of line AB is twice
the meridian distance of line AB
E
DMD of line EA is the
departure of line
Surveying - Traverse
DMD Calculations
N
Meridian distance
of line AB
A
E
B
Surveying - Traverse
DMD Calculations
 The DMD of any side is equal
to the DMD of the last side
plus the departure of the last
side plus the departure of the
present side
Side
AB
BC
CD
DE
EA
Balanced
Latitude Departure
-188.388
-152.253
29.933
139.080
171.627
-20.601
86.648
-195.470
-30.551
159.974
DMD
-20.601
 The DMD of line AB is departure of line AB
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse
DMD Calculations
Side
AB
BC
CD
DE
EA
-20.601 +
86.648 +
-195.470
-30.551
159.974
Side
DMD
-20.601
45.447
 The DMD of line BC is DMD of line AB + departure of
line AB + the departure of line BC
Surveying - Traverse
DMD Calculations
Side
AB
BC
CD
DE
EA
DMD
-20.601
-20.601
45.447
86.648
-195.470 + -63.375
-30.551 + -289.397
159.974
Surveying - Traverse
DMD Calculations
AB
BC
CD
DE
EA
-20.601
86.648 +
-195.470 +
-30.551
159.974
DMD
-20.601
45.447
-63.375
Surveying - Traverse
AB
BC
CD
DE
EA
Balanced
Latitude Departure
-188.388
-152.253
29.933
139.080
171.627
DMD
-20.601
-20.601
45.447
86.648
-63.375
-195.470
-30.551 + -289.397
159.974 + -159.974
 The DMD of line EA is DMD of line DE + departure of
line DE + the departure of line EA
Surveying - Traverse
Traverse Area - Double Area
 The sum of the products of each points DMD and
latitude equal twice the area, or the double area
Balanced
Latitude Departure
-188.388
-152.253
29.933
139.080
171.627
-188.388
-152.253
29.933
139.080
171.627
 The DMD of line CD is DMD of line BC + departure of
line BC + the departure of line CD
Side
 The DMD of line DE is DMD of line CD + departure of
line CD + the departure of line DE
Side
AB
BC
CD
DE
EA
Balanced
Latitude Departure
DMD Calculations
Balanced
Latitude Departure
-188.388
-152.253
29.933
139.080
171.627
Surveying - Traverse
DMD Calculations
Balanced
Latitude Departure
-188.388
-152.253
29.933
139.080
171.627
9/13
-20.601
86.648
-195.470
-30.551
159.974
DMD
-20.601
45.447
-63.375
-289.397
-159.974
 Notice that the DMD values can be positive or negative
Side
AB
BC
CD
DE
EA
Balanced
Latitude Departure
-188.388
-152.253
29.933
139.080
171.627
-20.601
86.648
-195.470
-30.551
159.974
DMD
Double Areas
-20.601
3,881
45.447
-63.375
-289.397
-159.974
 The double area for line AB equals DMD of line AB
times the latitude of line AB
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse
Surveying - Traverse
Traverse Area - Double Area
Traverse Area - Double Area
 The sum of the products of each points DMD and
latitude equal twice the area, or the double area
Side
AB
BC
CD
DE
EA
Balanced
Latitude Departure
-188.388
-152.253
29.933
139.080
171.627
-20.601
86.648
-195.470
-30.551
159.974
DMD
Double Areas
-20.601
3,881
-6,919
45.447
-63.375
-289.397
-159.974
Surveying - Traverse
Balanced
Latitude Departure
-20.601
86.648
-195.470
-30.551
159.974
Surveying - Traverse
Balanced
Latitude Departure
1 acre = 43,560 ft2
-20.601
86.648
-195.470
-30.551
159.974
 The sum of the products of each points DMD and
latitude equal twice the area, or the double area
AB
BC
CD
DE
EA
-188.388
-152.253
29.933
139.080
171.627
-20.601
86.648
-195.470
-30.551
159.974
DMD
Double Areas
-20.601
3,881
-6,919
45.447
-1,897
-63.375
-40,249
-289.397
-27,456
-159.974
 The double area for line EA equals DMD of line EA
times the latitude of line EA
 The sum of the products of each points DMD and
latitude equal twice the area, or the double area
Side
DMD
Double Areas
-20.601
3,881
45.447
-6,919
-63.375
-1,897
-289.397
-40,249
-159.974
-27,456
2 Area =
-72,641
Area =
Balanced
Latitude Departure
Traverse Area - Double Area
 The sum of the products of each points DMD and
latitude equal twice the area, or the double area
-188.388
-152.253
29.933
139.080
171.627
DMD
Double Areas
-20.601
3,881
-6,919
45.447
-1,897
-63.375
-289.397
-159.974
Surveying - Traverse
Traverse Area - Double Area
AB
BC
CD
DE
EA
-20.601
86.648
-195.470
-30.551
159.974
 The double area for line CD equals DMD of line CD
times the latitude of line CD
Side
DMD
Double Areas
-20.601
3,881
-6,919
45.447
-1,897
-63.375
-40,249
-289.397
-159.974
 The double area for line DE equals DMD of line DE
times the latitude of line DE
Side
-188.388
-152.253
29.933
139.080
171.627
Traverse Area - Double Area
 The sum of the products of each points DMD and
latitude equal twice the area, or the double area
-188.388
-152.253
29.933
139.080
171.627
AB
BC
CD
DE
EA
Balanced
Latitude Departure
Surveying - Traverse
Traverse Area - Double Area
AB
BC
CD
DE
EA
 The sum of the products of each points DMD and
latitude equal twice the area, or the double area
Side
 The double area for line BC equals DMD of line BC
times the latitude of line BC
Side
10/13
36,320 ft2
0.834 acre
AB
BC
CD
DE
EA
Balanced
Latitude Departure
-188.388
-152.253
29.933
139.080
171.627
1 acre = 43,560 ft2
-20.601
86.648
-195.470
-30.551
159.974
DMD
Double Areas
-20.601
3,881
45.447
-6,919
-63.375
-1,897
-289.397
-40,249
-159.974
-27,456
2 Area =
-72,641
Area =
36,320 ft2
0.834 acre
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse
Traverse Area - Double Area
11/13
Surveying - Traverse
Traverse Area - Double Area
 The word "acre" is derived from Old English æcer
(originally meaning "open field", cognate to Swedish
"åker", German acker, Latin ager and Greek αγρος
(agros).
 The word "acre" is derived from Old English æcer
(originally meaning "open field", cognate to Swedish
"åker", German acker, Latin ager and Greek αγρος
(agros).
 The acre was selected as approximately the amount of
land tillable by one man behind an ox in one day.
 A long narrow strip of land is more efficient to plough
than a square plot, since the plough does not have to be
turned so often.
 This explains one definition as the area of a rectangle
with sides of length one chain (66 ft) and one furlong
(ten chains or 660 ft).
 The word "furlong" itself derives from the fact that it
is one furrow long.
Surveying - Traverse
Traverse Area - Double Area
 The word "acre" is derived from Old English æcer
(originally meaning "open field", cognate to Swedish
"åker", German acker, Latin ager and Greek αγρος
(agros).
Surveying - Traverse
Traverse Area – Example 4
 Find the area enclosed by the following traverse
Side
Balanced
Latitude Departure
DMD
AB
BC
CD
DE
EA
600.0
100.0
0.0
-400.0
-300.0
Double Areas
200.0
400.0
100.0
-300.0
-400.0
2 Area =
1 acre = 43,560 ft2
Surveying - Traverse
DPD Calculations
 The same procedure used for DMD can be used the
double parallel distances (DPD) are multiplied by the
balanced departures
 The parallel distance of a line is the distance from
the midpoint of the line to the reference parallel or
east–west line
Area =
ft 2
acre
Surveying - Traverse
Rectangular Coordinates
 Rectangular coordinates are the convenient method
available for describing the horizontal position of
survey points
 With the application of computers, rectangular
coordinates are used frequently in engineering
projects
 In the US, the x–axis corresponds to the east–west
direction and the y–axis to the north–south direction
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse
Surveying - Traverse
Rectangular Coordinates Example
Rectangular Coordinates Example
 In this example, the length of AB is 300 ft and
bearing is shown in the figure below. Determine the
coordinates of point B
Latitude
y
B
Departure
N 42 30’ E
A
x
Coordinates of Point A
(200, 300)
AB
=300 ft cos(4230’)
= 221.183 ft
 In this example, it is assumed that the coordinates of
points A and B are know and we want to calculate the
latitude and departure for line AB
y
x B = 200 + 202.667 = 402.667 ft
B
y B = 300 + 221.183 = 521.183 ft
Side
AB
BC
CD
DE
EA
B
D
y
-20.601
86.648
-195.470
-30.551
159.974
x
C
Surveying - Traverse
Rectangular Coordinates Example
y
A
E
Side
AB
BC
CD
DE
EA
 y coordinates
= -400 ft
Departure
AB
= xB – xA
Departure
AB
= 220 ft
A
E
E = 0 ft
B
D
Side
AB
BC
CD
DE
EA
 x coordinates
x
C
Balanced
Latitude Depa rture
-188.388
-152.253
29.933
139.080
171.627
-20.601
86.648
-195.470
-30.551
159.974
A = E + 159.974 = 159.974 ft
B = A – 20.601 = 139.373 ft
C = B + 86.648 = 226.021 ft
D = C – 195.470 = 30.551 ft
E = D – 30.551 = 0 ft
Surveying - Traverse
Rectangular Coordinates Example
y
A (159.974, 340.640)
C = 0 ft
B
D
= yB – yA
AB
Coordinates of Point B
(320, -100)
Balanced
Latitude Depa rture
-188.388
-152.253
29.933
139.080
171.627
AB
Latitude
Rectangular Coordinates Example
 Consider our previous example, determine the x and y
coordinates of all the points
A
x
Latitude
Surveying - Traverse
Rectangular Coordinates Example
E
Coordinates of Point A
(100, 300)
A
AB =300 ft sin(4230’)
= 202.677 ft
Surveying - Traverse
y
12/13
C
x
Balanced
Latitude Depa rture
-188.388
-152.253
29.933
139.080
171.627
-20.601
86.648
-195.470
-30.551
159.974
D = C + 29.933 ft
E = D + 139.080 = 169.013 ft
A = E + 171.627 = 340.640 ft
B = A –188.388 = 152.252 ft
C = B –152.252 = 0 ft
B (139.373, 152.253)
(0.0, 169.013) E
(30.551, 29.933)
D
C (226.020, 0.0)
x
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse
Surveying - Traverse
Area Computed by Coordinates
Group Example Problem 5
 Compute the x and y coordinates from the following
balanced.
Balanced
Length (ft) Latitude Departure Latitude Departure
Bearing
Side
Points
Coordinates
x
y
degree m inutes
AB
BC
CD
DE
EA
S
S
N
N
N
6
29
81
12
42
15
38
18
24
59
W
E
W
W
E
189.53
175.18
197.78
142.39
234.58
939.46
-188.403
-152.268
29.916
139.068
171.607
-0.079
-20.634
86.617
-195.504
-30.576
159.933
-0.163
-188.388
-152.253
29.933
139.080
171.627
0.000
-20.601
86.648
-195.470
-30.551
159.974
0.000
A
B
C
D
E
100.000
100.000
Surveying - Traverse
Area Computed by Coordinates
y
A (159.974, 340.640)
Twice the area equals:
= 340.640(139.373 – 0.0)
B (139.373, 152.253)
(0.0, 169.013) E
+ 152.253(226.020 – 159.974)
C (226.020, 0.0) x
+ 29.933(0.0 – 226.020)
+ 169.013(159.974 – 30.551)
= 72,640.433 ft2
Area = 0.853 acre
(using a sign convention of + for next side and - for
last side)
Surveying - Traverse
Area Computed by Coordinates
 There is a simple variation of the coordinate
method for area computation
y
Area Computed by Coordinates
 There is a simple variation of the coordinate
method for area computation
A (159.974, 340.640)
B (139.373, 152.253)
(0.0, 169.013) E
Twice the area equals:
159.974(152.253) + 139.373(0.0) +
226.020(29.933) + 30.551(169.013) +
0.0(340.640)
- 340.640(139.373) – 152.253(226.020)
- 0.0(30.551) – 29.933(0.0)
– 169.013(159.974)
(30.551, 29.933) D
C (226.020, 0.0) x
A (159.974, 340.640)
x1
y1
= -72,640 ft2
Area = 36,320 ft2
x2
y2
x3
y3
x4
y4
B (139.373, 152.253)
(0.0, 169.013) E
x5
y5
x1
y1
Twice the area equals:
(30.551, 29.933) D
C (226.020, 0.0) x
Area = 36,320.2 ft2
Surveying - Traverse
y
 The area of a traverse can be computed by taking
each y coordinate multiplied by the difference in the
two adjacent x coordinates
+ 0.0(30.551 – 139.373)
(30.551, 29.933) D
13/13
= x1y2 + x2y3 + x3y4 + x4y5 + x5y1
- x2y1 – x3y2 – x4y3 – x5y4 – x1y5
End of Surveying - Traverse
Any Questions?